If cot θ = 7/8, evaluate: (i) (1-sin) (1+sinθ) / (1-cosθ) (1+cosθ) (ii) cot²θ

Given that: cot 0 = 7/8
Let cot 0=where k is a real number.
In △ABC, by Pythagoras theorem, we have
AC² = BC²+AB²
= (8k)²+(7k)²
= 64k²+49k²
= 113k²
⇒ AC = √(113k²) = √(113k)
(i) ((1+sin θ)(1-sin 9))/((1+cos θ)(1-cos θ))
= ((1+ 8/√113)(1- 8/√113))/((1+ 7√113)(1- 7/√113))
= (1-(8/√113)²)/(1-(7/√113)²) = (1- 64/113)/(1- 49/113) = ((113-64)/113)/((113-49)/113) = 49/64
(ii) cot² θ = (cot θ)² = (7/8)² = 49/64

Here is the Video explanation of the above question😀👇